Properties of Real Numbers

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A real number is one whose square is positive.

(https://docs.google.com/spreadsheets/d/1t6AckNp5MJ115DJ6Wo7kifqVGUkvzx6NEpSu_Fk1nQE/edit#gid=133220531) at #mathematics, on 20 September 02.30am UTC.

\[\huge{.999 \ldots = 1}\]

The above equation is a very common question that has misled many people over years. It even got a lengthy Wikipedia page. Is it true? They are two different numbers, no?

Surprisingly, it is true! You might have known this popular problem and hence memorized the answer. But do you know why this is true? More surprisingly, we can make this equation false! If we work on something other than the real numbers, that is. The above equation hinges on a very important property of the real numbers.

(problems)

Examples

Let \(x\in R\). Show that there exists an integer \(m\) such that \(m\le x < m+1\) and an integer \(l\) such that \(x<l\le x+1\).

Suppose that \(\alpha,\beta\) are two real numbers satisfying \(\alpha<\beta\). Show that there exist \(n_1,n_2 \in N\) such that \(\alpha<\alpha+\frac{1}{n_1}<\beta\) and \(\alpha<\beta-\frac{1}{n_2}<\beta.\)

\(0.999 \ldots\), being a number represented in decimal form, has the value \(\displaystyle 0 + \sum_{n=1}^\infty 9 \cdot \left( \frac{1}{10} \right)^n\). This is a consequence of any number that is represented in decimal: if the number is \(\overline{k.b_1 b_2 b_3 \ldots},\) where \(k\) is an integer and \(b_1, b_2, b_3, \ldots \in \{0,1,2,\ldots,9\},\) the value of any such number is \(\displaystyle k + \sum_{n=1}^\infty b_n \cdot \left( \frac{1}{10} \right)^n\) by definition. So we'd like to compute the value of \(\displaystyle S = \sum_{n=1}^\infty 9 \cdot \left( \frac{1}{10} \right)^n\).

A geometric sequence is a sequence of real numbers \(\{a_n\}_{n=1}^\infty\) that satisfies the property \(a_{n+1} = a_n \cdot r\) for some real number \(r\) and all positive integers \(n\). It can be shown by simple induction that if \(\{a_n\}\) forms a geometric sequence, there exist real numbers \(a\) and \(r\) such that \(a_n = ar^{n-1}\) for all positive integers \(n\). (Here we take \(0^0 = 1\).) A geometric series (or geometric progression) is a series whose elements form a geometric sequence, i.e. a series in the form \(\displaystyle \sum_{n=1}^\infty ar^{n-1}\). Observe that \(S\) is a geometric sequence with \(a = \frac{9}{10}, r = \frac{1}{10}\).

To compute the value of this series, we can take a look at its partial sums \(\displaystyle \sum_{n=1}^k ar^{n-1}\). This form is called a finite geometric series. Being the sum of a finite number of real numbers, we can apply the commutative and associative properties of addition and the distributive property of multiplication over addition, to manipulate the sum as such:

\[\displaystyle\begin{align*} S_k &= \sum_{n=1}^k ar^{n-1} \\ r S_k &= r \cdot \sum_{n=1}^k ar^{n-1} \\ &= \sum_{n=1}^k \left(r \cdot ar^{n-1}\right) \\ &= \sum_{n=1}^k ar^n \\ &= \sum_{n=2}^{k+1} ar^{n-1} \\ S_k - r S_k &= \left( \sum_{n=1}^k ar^{n-1} \right) - \left( \sum_{n=2}^{k+1} ar^{n-1} \right) \\ &= ar^0 + \sum_{n=2}^k \left(ar^{n-1} - ar^{n-1}\right) - ar^k \\ &= a\left(1-r^k\right) \\ S_k &= \frac{a\left(1-r^k\right)}{1-r} \\ &= \frac{a}{1-r} - \frac{a}{1-r} \cdot r^k. \end{align*}\]

By definition, the value of the series is simply the limit of its partial sums. Thus we'd like to compute \(\displaystyle \lim_{k \to \infty} S_k\), or \(\displaystyle \lim_{k \to \infty} \left( \frac{a}{1-r} - \frac{a}{1-r} \cdot r^k \right)\)

It's an easy exercise to show that if \(\{b_n\}\) is a sequence of real numbers that converges to a limit \(\displaystyle \lim_{n \to \infty} b_n\) and \(c\) is a constant, then \(\displaystyle \lim_{n \to \infty} (c + b_n) = c + \lim_{n \to \infty} b_n\) and \(\displaystyle \lim_{n \to \infty} (c \cdot b_n) = c \cdot \lim_{n \to \infty} b_n\). Since \(\frac{a}{1-r}\) and \(-\frac{a}{1-r}\) are constants, applying these properties gives:

\[\displaystyle\begin{align*} \lim_{k \to \infty} S_k &= \lim_{k \to \infty} \left( \frac{a}{1-r} - \frac{a}{1-r} \cdot r^k \right) \\ &= \frac{a}{1-r} + \lim_{k \to \infty} \left( \left(- \frac{a}{1-r} \right) \cdot r^k \right) \\ &= \frac{a}{1-r} + \left(- \frac{a}{1-r} \right) \cdot \lim_{k \to \infty} r^k. \end{align*}\]

Our problem has been reduced to determining \(\displaystyle \lim_{k \to \infty} r^k\).

For the purpose of proving \(0.999 \ldots = 1\), it suffices to restrict our attention to \(0 < r < 1\). It is simple induction again to show that \(0 < r^{k+1} < r^k\) for all positive integers \(k\). From the fact that \(r^k > r^{k+1}\), we know that the sequence \(\{r^k\}\) is monotonically decreasing, and thus converges to a limit \(L\) by monotone convergence theorem. From the fact that \(r^k > 0\), \(L\) cannot be less than \(0\); if \(L < 0\), then simply take \(\epsilon = -L\), and now there is no element of \(\{r^k\}\) that lies in \((L-\epsilon, L+\epsilon)\) (because \(L+\epsilon = 0 < r^k\)). Hence there can be no \(N\) where \(r^k \in (L-\epsilon, L+\epsilon)\) for all \(k > N\) (the condition necessary for a convergent sequence to converge to the limit).

todo: show that \(L > 0\) is impossible

Thus we have shown that \(L < 0\) and \(L > 0\) lead to a contradiction. The Archimedean property tells us that \(L\) cannot be infinitesimal either, since there is no non-zero infinitesimal in the real numbers. The remaining option is thus \(L = 0\).

Hence, \(\displaystyle \lim_{k \to \infty} S_k = \frac{a}{1-r} + \left(- \frac{a}{1-r} \right) \cdot 0\), or \(\displaystyle \lim_{k \to \infty} S_k = \frac{a}{1-r}\); the geometric series, for \(0 < r < 1\), gives the value \(\frac{a}{1-r}\).

But \(S\) is a geometric series, with \(a = \frac{9}{10}, r = \frac{1}{10}\). Since \(0 < r < 1\), we can apply the result we obtained above to get \(S = \frac{9/10}{1 - 1/10} = 1\). This proves that \(0.999 \ldots = 1\).

Density of Q in R